Simplify the following expression: $y = \dfrac{-5x^2+3x+2}{x - 1}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-5)}{(2)} &=& -10 \\ {a} + {b} &=& &=& {3} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-10$ and add them together. Remember, since $-10$ is negative, one of the factors must be negative. The factors that add up to ${3}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-2}$ and ${b}$ is ${5}$ $ \begin{eqnarray} {ab} &=& ({-2})({5}) &=& -10 \\ {a} + {b} &=& {-2} + {5} &=& 3 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-5}x^2 {-2}x) + ({5}x +{2}) $ Factor out the common factors: $ x(-5x - 2) - 1(-5x - 2)$ Now factor out $(-5x - 2)$ $ (-5x - 2)(x - 1)$ The original expression can therefore be written: $ \dfrac{(-5x - 2)(x - 1)}{x - 1}$ We are dividing by $x - 1$ , so $x - 1 \neq 0$ Therefore, $x \neq 1$ This leaves us with $-5x - 2; x \neq 1$.